Answer:
F = 3.86 x 10⁻⁶ N
Explanation:
First, we will find the distance between the two particles:
where,
r = distance between the particles = ?
(x₁, y₁, z₁) = (2, 5, 1)
(x₂, y₂, z₂) = (3, 2, 3)
Therefore,
Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:
where,
F = magnitude of force = ?
k = Coulomb's Constant = 9 x 10⁹ Nm²/C²
q₁ = magnitude of first charge = 2 x 10⁻⁸ C
q₂ = magnitude of second charge = 3 x 10⁻⁷ C
r = distance between the charges = 3.741 m
Therefore,
<u>F = 3.86 x 10⁻⁶ N</u>
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
A transverse wave is a wave where the particles in the medium move perpendicular (at right angles) to the direction of the source or its propagation (think of a snake slithering through grass) an example of a transverse wave could be a light wave. Light waves for instance don’t need a medium in order to propagate but transverse waves in general do need a medium.
They typically represents different wavelengths of element due to its energy emission in the form of visible light. When an electron of that particular element move from a higher energy level down to a lower energy level, it gives off energy in the form of photon emission. Atom of a certain element has a unique electron arrangement thus it can considered that particular element's spectrum is unique.
Answer:
The handrails must be approximately 10.63 meters long
Explanation:
The given parameters are;
The height of the bleachers, h = 8 m
The depth of the bleachers, d = 7 m
The length of the hand rails to go along the bleachers from bottom to top is given by Pythagoras' Theorem as follows;
The length of the hand rail = √(d² + h²)
∴ The length of the hand rail = √(7² + 8²) = √113 ≈ 10.63
In order for the handrails to go along the bleachers from top to bottom, they must be approximately 10.63 meters long.
